N-th Tribonacci Number

class Solution {
public:
    int tribonacci(int n) {
        if (n < 3) {
            return n == 0 ? 0 : 1;
        }
            
        vector<int> trib(n + 1, 0);
        trib[1] = 1;
        trib[2] = 1;
            
        for (int i = 3; i <= n; ++i) {
            trib[i] = trib[i - 1] + trib[i - 2] + trib[i - 3];
        }
            
        return trib[n];
    }
};

The provided code snippet is a C++ solution designed to calculate the N-th number in the Tribonacci sequence. The Tribonacci sequence is a generalization of the Fibonacci sequence where each number is the sum of the three preceding ones.

Here’s a concise explanation of the approach and method used in the solution:

• Check for base cases:
  • Return 0 if n is 0.
  • Return 1 if n is 1 or 2.
• Initialize a vector, trib, that supports up to n elements and set default values to 0.
• Set initial known values:
  • trib[1] and trib[2] to 1.
• Iterate from 3 to n:
  • Update trib[i] to be the sum of the three preceding elements: trib[i-1], trib[i-2], and trib[i-3].
• Finally, return trib[n], which contains the N-th Tribonacci number.

This solution effectively utilizes dynamic programming to build up to the required N-th number by storing previously computed values, thus avoiding the exponential time complexity of a naive recursive approach. The utilization of a vector ensures that each number is computed only once.

class Solution {
    public int tribonacciNumber(int num) {
        if (num < 3) {
            return num == 0 ? 0 : 1;
        }
            
        int[] sequence = new int[num + 1];
        sequence[1] = 1;
        sequence[2] = 1;
            
        for (int index = 3; index <= num; ++index) {
            sequence[index] = sequence[index - 1] + sequence[index - 2] + sequence[index - 3];
        }
            
        return sequence[num];
    }
}

This Java solution efficiently computes the N-th tribonacci number using dynamic programming. The tribonacci sequence, like the Fibonacci sequence, starts with certain fixed numbers and each subsequent number is the sum of the three preceding numbers. The method tribonacciNumber(int num) determines this N-th number in the sequence.

• The method starts by handling the base cases:
  • If num is less than 3, the method returns 0 for num == 0 and 1 for num == 1 or 2.
• It initializes an array sequence to store computed tribonacci numbers, thereby avoiding redundant calculations with each function call.
• The array is pre-populated with values for sequence[1] and sequence[2] both set to 1, considering the sequence definition.
• A for-loop then iterates from 3 to num, computing the tribonacci number at each position by summing the three immediately preceding numbers in the sequence.
• Finally, the method returns the value at sequence[num], which contains the N-th tribonacci number.

This approach leverages the dynamic programming principle of overlapping subproblems, ensuring the calculation is efficient and optimal. Memory usage is linear in terms of the input size num, making it a scalable solution. With this solution, compute any desired N-th number in the tribonacci sequence by passing the appropriate number to the tribonacciNumber method.

class Result:
    def trib(self, N: int) -> int:
        if N < 3:
            return 1 if N else 0
        result = [0] * (N + 1)
        result[1], result[2] = 1, 1
        for index in range(3, N + 1):
            result[index] = result[index - 1] + result[index - 2] + result[index - 3]
        return result[N]

The provided Python function calculates the N-th number in the Tribonacci sequence, which is a variation of the Fibonacci sequence where each number is the sum of the three preceding ones. This program, encapsulated within a class Result, includes a method trib that determines the Tribonacci value at position N.

• The method starts by managing base cases: if N is less than 3, it returns 1 for N equals 1 or 2, and returns 0 for N equals 0.
• To store the Tribonacci numbers, the function initializes a list called result with a size of N + 1, prefilling the values of result[1], and result[2] to 1 since the first two nonzero elements of the Tribonacci series are both 1.
• A loop then computes each subsequent Tribonacci number from the 3rd to the N-th by summing the three previous numbers in the sequence.
• The function returns the value result[N] to output the N-th Tribonacci number.

Optimized for readability and efficiency, the function prepares for large values of N by using a dynamic programming approach that stores previously calculated values in an array. This avoids redundant calculations, making the method scalable and fast for larger inputs.